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A force of 6.0 N acts on a charge of 3.0 mu C when it is placed in a uniform ele

ID: 2076559 • Letter: A

Question

A force of 6.0 N acts on a charge of 3.0 mu C when it is placed in a uniform electric field. What is the magnitude of this electric field? A) 2.0 MN/C B) 18 MN/C C) 0.50 MN/C D) 1.0 MN/C E)_ 130 MN/C My flashlight beam makes an angle of 60 degrees with the surface of the water before it enters the water. In the water what angle does the beam make with the surface? A) 22 degree B) 30 degree C) 60 degree D) 0 degree E) 68 degree Two point charges mu C and -8.00 mu C are separated by a distance of 20.0 cm. A + 7.00 mu C charge is placed midway between these two charges. What is the electric force acting on this charge of the other two charges? A). 1.76 N directed towards the negative charge B) 0.453 N directed towards the negative charge C) 176 N directed towards the negative charge D) 1.76 N directed towards the positive charge E) 176 N directed towards the positive charge In a two-slit experiment, the slit separation is 2.60 times 10^-5 m. The interference pattern is created on a screen that is 2.00 m away from the slits, If the 7^th bright fringe on the screen is a linear distance of 10.00 cm away from the central fringe. what is the wavelength of the light? A) 100 nm B) 1958 nm C) 205 nm D) 175 nm E) 185 nm In an RLC circuit, the resistance is 10 ohms, the capacitive reactance is 30. ohms, and the inductive reactance is 50. ohms. What is the circuit impedance? A) 63 Ohm B) 22. Ohm C) 81. Ohm D) 10.Ohm E) 90.Ohm The charge carried by one electron is e = -1.6 times 10^-19 C. The number of excess electrons necessary to produce a charge of 1.0 C is A) 6.3 times 10^6 B) 1.6 times 10^19 C) 6.3 times 10^18 D) 1.6 times 10^18 E) 6.3 times 10^19

Explanation / Answer

67. n1sin(theta1) = n2sin(theta2) - Snell's Law
The angle theta is measured between the beam of light and the NORMAL where
n1 = 1
theta1 = 30 degrees
n2 = 1.33

Put all values in equation, thetha = 22°

70. impedence

Z=sqrt[(R^2+(XL^2-Xc^2)]

=sqrt[100+400]=10sqrt(5)=22.36 ohms

71. E = 1.6 *10^-19c

1e /1.6 * 10^-19

1/ 1e/1.6 ^103

6.3 * 10^18c

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